Re: Sets question (Proving)

Consider what intersections you get when you apply the first distributive law that Zarathustra gave above to the left-hand side

(A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A).

Each resulting intersection will have one set from each of the three sets of parentheses. Thus, there will be

A ∩ B ∩ C,
A ∩ B ∩ A,
A ∩ C ∩ C,
A ∩ C ∩ A,

and so on. Note that repetitions can be removed because X ∩ X = X for any set X.

Each intersection will have either two sets (A ∩ B or A ∩ C or B ∩ C) or all three sets (A ∩ B ∩ C). Also, each of A ∩ B, A ∩ C, and B ∩ C will be a part of the left-hand side. Finally, note that A ∩ B ∩ C ⊆ A ∩ B, and if X ⊆ Y, then X ∪ Y = Y, so A ∩ B ∩ C can be omitted from the final union.